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Singly Excited Magnetic System - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Singly Excited Magnetic System (10 Questions)

You can prepare effectively for Electrical Engineering (EE) Electrical Machines with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Singly Excited Magnetic System". These 10 questions have been designed by the experts with the latest curriculum of Electrical Engineering (EE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 10

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Test: Singly Excited Magnetic System - Question 1

For a toroid to extract the energy from the supply system, the flux linkages of the magnetic field must be ________

Detailed Solution: Question 1

dWelec = idφ = eidt, where dWelec = differential electrical energy to coupling field, and if the flux linkages are either constant or zero, i.e, dφ = 0, then dWelec = 0.

Test: Singly Excited Magnetic System - Question 2

The energy stored in a magnetic field is given by ____________ where L = self-inductance and Rl=reluctance.

Detailed Solution: Question 2

We know that Wfld = 1/2 φi and L = φ/i, thus Wfld = 1/2 Li2.

Test: Singly Excited Magnetic System - Question 3

For a linear electromagnetic circuit, which of the following statement is true?

Detailed Solution: Question 3

Wfld = Wfld= 1/2 φ*i = 1/2 F*∅.

Test: Singly Excited Magnetic System - Question 4

The electromagnetic force developed in any physical system acts in such a direction as to tend to _____________

Detailed Solution: Question 4

fe = (∂Wfld1 (i,x))/∂x = (∂Wfld1 (F,x))/∂x, the positive sign before ∂Wfld1 indicates that force fe acts in a direction as to tend to increase the co-energy at constant mmf.

Test: Singly Excited Magnetic System - Question 5

The electromagnetic torque developed in any physical system, and with magnetic saturation neglected, acts in such a direction as to tend to ____________

Detailed Solution: Question 5

fe = 1/2 ∅2 dRl/dx, Te = -1/2 ∅2 dRl/dθ = 1/2 i2 dL/dθ.

Test: Singly Excited Magnetic System - Question 6

Magnetic stored energy density for iron is given by ______

Detailed Solution: Question 6

Magnetic stored energy density for iron is given as
wfld = Wfld/((Length of the magnetic path through Iron)*(Iron area normal to the magnetic flux)) = 1/2 (F∅)/(length*Area) = 1/2 F/length ∅/area = 1/2 H*B
Also, H = B/μ,thus wfld = 1/2 B2/μ.

Test: Singly Excited Magnetic System - Question 7

When a current of 5A flows through a coil of linear magnetic circuit, it has flux linkages of 2.4 wb-turns. What is the energy stored in the magnetic field of this coil in Joules?

Detailed Solution: Question 7

Wfld = 1/2 φ*i = 1/2*2.4*5 = 6 Joules.

Test: Singly Excited Magnetic System - Question 8

The electromagnetic force and/or torque, developed in any physical system, acts in such a direction as to tend to ____________

Detailed Solution: Question 8

fe = -(∂Wfld (φ, x))/∂x = -(∂Wfld (∅,x))/∂x and Te = -(∂Wfld(φ,θ))/∂θ = -(∂Wfld (∅,θ))/∂θ
The negative sign before ∂Wfld indicates that fe acts in a direction as to tend to decrease the stored energy at constant mmf.

Test: Singly Excited Magnetic System - Question 9

Consider a magnetic relay with linear magnetization curve in both of its open and closed position. What happens to the electrical energy input to the relay, when the armature moves slowly from open position to closed position?

Detailed Solution: Question 9

For the above mentioned case, Wfld = Wmech and Wfld = Welec/2 hence, option “c” is the correct answer.

Test: Singly Excited Magnetic System - Question 10

Electromagnetic force and/or torque developed in any physical system, acts in such a direction as to tend to ____________

Detailed Solution: Question 10

fe = (∂Wfld1(i,x))/∂x = (∂Wfld(i,x))/∂x.

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